Step 1:
Given :
$\cos^{-1}\sqrt p+\cos^{-1}\sqrt{1-p}+\cos^{-1}\sqrt{1-q}=\large\frac{3\pi}{4}$
$\Rightarrow \cos^{-1}\sqrt p+\sin^{-1}\sqrt p+\cos^{-1}\sqrt{1-q}=\large\frac{3\pi}{4}$
Let $\cos^{-1}\sqrt{1-p}=\theta$
$\cos\theta=\sqrt{1-p}$
$\sin\theta=\sqrt p$
$\sin^{-1}x+\cos^{-1}x=\large\frac{\pi}{2}$
$\large\frac{\pi}{2}$$+\cos^{-1}\sqrt{1-q}=\large\frac{3\pi}{4}$
$\cos^{-1}\sqrt{1-q}=\large\frac{\pi}{4}$
$\sqrt{1-q}=\cos\large\frac{\pi}{4}$
$\Rightarrow \large\frac{1}{\sqrt 2}$
$1-q=\large\frac{1}{2}$
$q=\large\frac{1}{2}$
Step 2:
$\cos^{-1}\sqrt p\Rightarrow 0\leq \sqrt p\leq 1$
$\Rightarrow 0\leq p\leq 1$
$\cos^{-1}\sqrt{1-p}=0\leq \sqrt{1-p}\leq 1$
$\Rightarrow 0\leq p\leq 1$
$\therefore 0\leq p\leq 1$ and $q=\large\frac{1}{2}$
Hence (c) is the correct option.